The area under the curve #y=1/x# from #x=8# to #x=10# is revolved about the #x#-axis. What is the volume of the solid formed?

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Answer 1 · 2017-04-01T06:28:16

#pi/40#

The region bound by these curves can be thought of as the region bound by #y=1/x# and the #x# axis, between #x=8# and #x=10#.

Hence, #V=piint_8^10(1/x)^2dx#

#=piint_8^10(1/(x^2))dx#

#=piint_8^10x^-2dx#

#=pi[x^-1/-1]_8^10#

#=-pi[1/x]_8^10#

#=-pi(1/10-1/8) #

#=-pi(-1/40) #
#=pi/40#